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Question:

The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the wires has a length of 1 m at 10℃. Now the end P is maintained at 10℃, while the end S is heated and maintained at 400℃. The system is thermally insulated from its surroundings. If the thermal conductivity of wire PQ is twice RS and the coefficient of linear thermal expansion of PQ is 1.2 × 10⁻⁵ K⁻¹, the change in length of the wire PQ is:

1.56 mm

0.78 mm

2.34 mm

0.90 mm

Solution:

Let the temperature of the junction be T.
∴rate of heat transfer = dQ/dt = 2KA(T - 10)/L = KA(400 - T)/L
⇒ 2(T - 10) = 400 - T
or, T = 140 ℃
Now, for the wire PQ, let us imagine a small length Δx at a distance x from the junction.
∴ ΔT/Δx = (140 - 10)/1 = 130
So, temperature at distance x: T = 10 + 130x
or, T - 10 = 130x
Increase in length of the small element Δx is expressed as:
dy/dx = αΔT = α(T - 10)
or, dy/dx = α × 130x
Integrating both sides, we get:
∫₀ΔL dy = 130α∫₀¹ x dx
∵ L = 1 m (Given)
or, ΔL = 130αx²/2 = 130 × 1.2 × 10⁻⁵ × 1²/2 = 0.78 × 10⁻³ m = 0.78 mm